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      <title>如何判断链表有环 - 交互式测验</title>
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      <div class="main-wrapper">
          <div class="quiz-container">
            <div class="quiz-header">
              <h1>如何判断链表有环 - 交互式测验</h1>
              <p>检验你对文档内容的理解程度</p>
            </div>
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      <script>
        const quizData = [
  {
    "question": "根据文档，快慢指针法（Floyd判圈算法）是如何判断链表中存在环的？",
    "options": [
      "一个指针从头开始，另一个从尾开始，如果它们相遇则有环。",
      "一个快指针每次移动两步，一个慢指针每次移动一步，如果它们最终相遇，则说明存在环。",
      "两个指针以相同的速度移动，如果其中一个指针回到起点，则说明有环。",
      "快指针每次移动三步，慢指针每次移动一步，以更快地检测到环。"
    ],
    "correctAnswerIndex": 1,
    "explanation": "快慢指针法的核心在于速度差。在一个有环的链表中，快指针（每次移动两步）最终必然会从后面追上慢指针（每次移动一步），这个追及相遇的现象证明了环的存在。这就像在圆形跑道上，跑得快的人总会套圈跑得慢的人。",
    "source": "文档中「快慢指针法」部分",
    "hint": "文档描述该方法利用了两个指针在链表中的...速度不同的特点来检测环。"
  },
  {
    "question": "在使用快慢指针法时，如果链表中不存在环，算法会通过什么条件来终止？",
    "options": [
      "慢指针到达链表的末尾（指向 null）。",
      "算法会陷入无限循环，无法终止。",
      "快指针到达链表的末尾（其自身或其下一个节点指向 null）。",
      "两个指针的移动步数超过了预设的阈值。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "在一个没有环的有限链表中，快指针由于移动速度更快，它必然会首先到达链表的终点。当快指针自身为 `null` 或它的 `next` 指针为 `null` 时，它将无法再继续移动两步，此时循环终止，可以断定链表无环。",
    "source": "文档中「快慢指针法」部分",
    "hint": "思考一下，在一条直路上，谁会先到达终点？文档中提到：\"如果快指针走到了链表的尽头...\""
  },
  {
    "question": "文档中描述的哈希表法，其检测链表环的核心原理是什么？",
    "options": [
      "将链表所有节点的值（val）存入哈希表，如果发现有重复的值，则说明有环。",
      "遍历链表，将每个访问过的节点对象本身存入哈希表，如果遇到一个节点在哈希表中已存在，则证明有环。",
      "使用哈希表计算每个节点的哈希值，如果哈希值发生碰撞，则说明有环。",
      "将链表分为两半，对两半分别计算哈希值，如果哈希值相等则有环。"
    ],
    "correctAnswerIndex": 1,
    "explanation": "哈希表法利用其近乎O(1)的查找效率来记录访问历史。它存储的是节点对象（或其内存地址/引用），而非节点的值。当遍历到一个新节点时，先检查它是否已在哈希表中。如果在，说明这个节点之前被访问过，从而形成了一个闭环。",
    "source": "文档中「哈希表法」部分",
    "hint": "文档提到：\"每遍历到一个新节点，就用新节点和HashSet集合当中存储的节点作比较，如果发现HashSet当中存在相同节点ID...\""
  },
  {
    "question": "根据文档，节点标记法被认为“不太推荐”的主要原因是什么？",
    "options": [
      "它的时间复杂度远高于其他方法。",
      "它需要修改原始链表的结构或数据。",
      "它的空间复杂度非常高。",
      "它只能处理值为正整数的链表。"
    ],
    "correctAnswerIndex": 1,
    "explanation": "文档明确指出：“这种方法需要修改原始链表的结构，不太推荐。” 在许多编程场景和算法竞赛中，修改输入数据被认为是一种不好的实践，因为它可能会引入意料之外的副作用，影响程序的其他部分或后续处理。",
    "source": "文档中「节点标记法」部分",
    "hint": "在「节点标记法」一节的结尾，文档给出了一个直接的评价和原因。"
  },
  {
    "question": "在使用快慢指针法找到了环内的相遇点后，文档描述的找到“环的入口节点”的后续步骤是什么？",
    "options": [
      "相遇点就是环的入口节点。",
      "将慢指针重置到链表头，快指针不动，然后两个指针同步移动，再次相遇点即为入口。",
      "先确定环的长度，然后将一个指针从头节点先移动环的长度步，再让两个指针以相同速度同步移动，它们的相遇点即为入口。",
      "从相遇点开始，让慢指针继续移动，直到它第二次回到相遇点，记录路径上的第一个节点即为入口。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "文档描述了一个基于环长度的精确方法。在快慢指针相遇后，首先通过额外遍历计算出环的长度 L。然后，将两个指针都放回链表头，让其中一个指针先前进 L 步。之后，两个指针再以相同的速度（每次一步）一起前进，它们再次相遇的地方就是环的入口。",
    "source": "文档中「如何找到环的入口」部分",
    "hint": "文档中的第三个步骤描述了如何利用已知的环长度来定位入口。关键在于让一个指针先走一段距离。"
  },
  {
    "question": "对比快慢指针法和哈希表法，以下关于它们空间复杂度的描述哪项是正确的？",
    "options": [
      "两者的空间复杂度都是O(1)。",
      "快慢指针法的空间复杂度是O(n)，哈希表法是O(1)。",
      "快慢指针法的空间复杂度是O(1)，而哈希表法是O(n)。",
      "两者的空间复杂度都是O(n)。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "快慢指针法只需要两个额外的指针变量来存储快慢指针的引用，其所需空间是固定的，不随链表长度n的变化而变化，因此空间复杂度为O(1)。哈希表法需要一个哈希集合来存储所有访问过的节点，在最坏情况下（链表无环），需要存储n个节点，因此空间复杂度为O(n)。",
    "source": "文档中「哈希表法」部分",
    "hint": "哈希表法的描述中提到：\"这种方法的空间复杂度较高...\"，而快慢指针法只需创建两个对象引用。"
  }
];
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            const wrongCount = answeredCount - correctCount;
            const skippedCount = totalQuestions - answeredCount;
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                '<div class="score-card">' +
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                '<div class="score-card">' +
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        function reviewQuiz() {
            const reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }

            document.querySelector('.quiz-body').style.display = 'block';
            document.getElementById('quiz-footer').style.display = 'flex';
            
            currentQuestionIndex = 0;
            loadQuestion();
            updateNavigationButtons(); // Ensure nav buttons are correctly displayed
        }

        // Event Delegation for options
        optionsListEl.addEventListener('click', (event) => {
            if (event.target && event.target.matches('li.option-item')) {
                // Check if the question has already been answered
                if (userAnswers[currentQuestionIndex] === null) {
                    const selectedIndex = parseInt(event.target.dataset.index, 10);
                    selectOption(selectedIndex);
                }
            }
        });

        prevBtn.addEventListener('click', goToPreviousQuestion);
        nextBtn.addEventListener('click', goToNextQuestion);
        hintBtn.addEventListener('click', showHint);
        
        // 初始加载
        loadQuestion();
      </script>
    </body>
    </html>
  